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Question Number 168278 by Florian last updated on 07/Apr/22

((log_3 (12))/(log_(36) (3)))−((log_3 (4))/(log_(108) (3))) = x x =

$$\:\:\:\:\:\frac{{log}_{\mathrm{3}} \left(\mathrm{12}\right)}{{log}_{\mathrm{36}} \left(\mathrm{3}\right)}−\frac{{log}_{\mathrm{3}} \left(\mathrm{4}\right)}{{log}_{\mathrm{108}} \left(\mathrm{3}\right)}\:=\:{x} \\ $$$$\:\:\:\:\:{x}\:=\: \\ $$

Commented by benhamimed last updated on 07/Apr/22

(((ln 12)/(ln 3))/((ln 3)/(ln 36)))−(((ln 4)/(ln 3))/((ln 3)/(ln 108)))=x ((ln 12×ln 36−ln 4×ln 108)/(ln^2 3))=x (((2ln 2+ln 3)2(ln 2+ln 3)−2ln 2(2ln2+3ln 3) )/(ln^2 3))=x ((2(ln 2+ln 3)ln 3−2ln 2ln 3)/(ln ^2 3))=x ((2ln 3)/(ln 3))=x x=2

$$\frac{\frac{\mathrm{ln}\:\mathrm{12}}{\mathrm{ln}\:\mathrm{3}}}{\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{ln}\:\mathrm{36}}}−\frac{\frac{\mathrm{ln}\:\mathrm{4}}{\mathrm{ln}\:\mathrm{3}}}{\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{ln}\:\mathrm{108}}}={x} \\ $$$$\frac{\mathrm{ln}\:\mathrm{12}×\mathrm{ln}\:\mathrm{36}−\mathrm{ln}\:\mathrm{4}×\mathrm{ln}\:\mathrm{108}}{\mathrm{ln}\:^{\mathrm{2}} \mathrm{3}}={x} \\ $$$$\frac{\left(\mathrm{2ln}\:\mathrm{2}+\mathrm{ln}\:\mathrm{3}\right)\mathrm{2}\left(\mathrm{ln}\:\mathrm{2}+\mathrm{ln}\:\mathrm{3}\right)−\mathrm{2ln}\:\mathrm{2}\left(\mathrm{2ln2}+\mathrm{3ln}\:\mathrm{3}\right)\:}{\mathrm{ln}\:^{\mathrm{2}} \mathrm{3}}={x} \\ $$$$\frac{\mathrm{2}\left(\mathrm{ln}\:\mathrm{2}+\mathrm{ln}\:\mathrm{3}\right)\mathrm{ln}\:\mathrm{3}−\mathrm{2ln}\:\mathrm{2ln}\:\mathrm{3}}{\mathrm{ln}\overset{\mathrm{2}} {\:}\mathrm{3}}={x} \\ $$$$\frac{\mathrm{2ln}\:\mathrm{3}}{\mathrm{ln}\:\mathrm{3}}={x} \\ $$$${x}=\mathrm{2} \\ $$

Commented by Florian last updated on 07/Apr/22

Very Good !

$${Very}\:{Good}\:! \\ $$

Answered by greogoury55 last updated on 07/Apr/22

⇒x = log _3 (9×4).log _3 (3×4)−log _3 (4).log _3 (27×4) ⇒x=(2+log _3 (4))(1+log _3 (4))−log _3 (4)(3+log _3 (4)) let log _3 (4)= d ⇒x=(2+d)(1+d)−d(3+d) ⇒x=2+3d+d^2 −3d−d^2 ⇒x=2

$$\Rightarrow{x}\:=\:\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{9}×\mathrm{4}\right).\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{3}×\mathrm{4}\right)−\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{4}\right).\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{27}×\mathrm{4}\right) \\ $$$$\Rightarrow{x}=\left(\mathrm{2}+\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{4}\right)\right)\left(\mathrm{1}+\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{4}\right)\right)−\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{4}\right)\left(\mathrm{3}+\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{4}\right)\right) \\ $$$${let}\:\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{4}\right)=\:{d} \\ $$$$\Rightarrow{x}=\left(\mathrm{2}+{d}\right)\left(\mathrm{1}+{d}\right)−{d}\left(\mathrm{3}+{d}\right) \\ $$$$\Rightarrow{x}=\mathrm{2}+\mathrm{3}{d}+{d}^{\mathrm{2}} −\mathrm{3}{d}−{d}^{\mathrm{2}} \\ $$$$\Rightarrow{x}=\mathrm{2} \\ $$

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